Physics (School & Undergraduate) MCQ & Objective Questions
Physics is a fundamental subject that plays a crucial role in school and undergraduate exams. Mastering Physics concepts not only enhances your understanding of the universe but also significantly boosts your exam scores. Practicing MCQs and objective questions helps you identify important topics and improves your problem-solving skills, making it an essential part of your exam preparation.
What You Will Practise Here
Newton's Laws of Motion and their applications
Work, Energy, and Power concepts and formulas
Waves and Sound: Properties and equations
Optics: Reflection, refraction, and lens formulas
Thermodynamics: Laws and key definitions
Electromagnetism: Basics of electric fields and circuits
Modern Physics: Introduction to quantum mechanics and relativity
Exam Relevance
Physics is a significant part of the curriculum for CBSE, State Boards, NEET, and JEE exams. Questions often focus on conceptual understanding and application of formulas. Common patterns include numerical problems, theoretical questions, and diagram-based queries. Familiarizing yourself with these patterns through practice is vital for success in these competitive exams.
Common Mistakes Students Make
Misunderstanding the application of Newton's Laws in different scenarios
Confusing work done with energy concepts
Overlooking the importance of units and dimensions in calculations
Neglecting to draw diagrams for problems related to optics and mechanics
Failing to relate theoretical concepts to practical examples
FAQs
Question: What are some effective ways to prepare for Physics MCQs?Answer: Regular practice of MCQs, understanding key concepts, and revising important formulas are effective strategies for preparation.
Question: How can I improve my problem-solving speed in Physics exams?Answer: Practice timed quizzes and focus on solving a variety of problems to enhance your speed and accuracy.
Don't wait any longer! Start solving practice MCQs today to test your understanding and boost your confidence in Physics. Remember, consistent practice is the key to mastering important Physics (School & Undergraduate) questions for exams.
Q. If two point charges, +Q and -Q, are separated by a distance d, what is the magnitude of the electric field (E) at the midpoint?
A.
E = 0
B.
E = k * Q / (d/2)^2
C.
E = k * Q / d^2
D.
E = k * Q / (d^2/4)
Show solution
Solution
At the midpoint, the electric fields due to both charges cancel each other out, resulting in E = 0.
Correct Answer:
A
— E = 0
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Q. If two resistors, R1 and R2, are in series, what is the equivalent resistance (R_eq)?
A.
R_eq = R1 + R2
B.
R_eq = R1 * R2
C.
R_eq = R1 / R2
D.
R_eq = R1 - R2
Show solution
Solution
The equivalent resistance of resistors in series is given by R_eq = R1 + R2.
Correct Answer:
A
— R_eq = R1 + R2
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Q. In a Carnot engine operating between 300 K and 600 K, what is the maximum efficiency?
A.
50%
B.
60%
C.
33.33%
D.
100%
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Solution
Efficiency = 1 - (T_cold/T_hot) = 1 - (300/600) = 0.5 or 50%.
Correct Answer:
B
— 60%
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Q. In a Carnot engine operating between 500 K and 300 K, what is the maximum efficiency?
A.
40%
B.
60%
C.
50%
D.
20%
Show solution
Solution
Efficiency = 1 - (Tc/Th) = 1 - (300/500) = 0.4 or 40%.
Correct Answer:
B
— 60%
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Q. In a Carnot engine, if the hot reservoir is at 500 K and the cold reservoir is at 300 K, what is the maximum efficiency?
A.
40%
B.
60%
C.
50%
D.
20%
Show solution
Solution
The maximum efficiency of a Carnot engine is given by 1 - (T_c/T_h), which is 1 - (300/500) = 0.4 or 40%.
Correct Answer:
B
— 60%
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Q. In a circuit with a 12 V battery and a 4 Ω resistor, what is the current flowing through the circuit?
A.
3 A
B.
2 A
C.
4 A
D.
1 A
Show solution
Solution
Using Ohm's law, I = V/R = 12 V / 4 Ω = 3 A.
Correct Answer:
B
— 2 A
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Q. In a circuit with a 12 V battery and a resistor of 4 Ω, what is the current flowing through the circuit?
A.
3 A
B.
4 A
C.
2 A
D.
6 A
Show solution
Solution
Using Ohm's Law: I = V/R = 12 V / 4 Ω = 3 A.
Correct Answer:
A
— 3 A
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Q. In a circuit with a 12V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?
Show solution
Solution
Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.
Correct Answer:
B
— 6V
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Q. In a circuit with a 24V battery and a total resistance of 6Ω, what is the total current flowing through the circuit?
Show solution
Solution
Using Ohm's Law, I = V / R = 24V / 6Ω = 4A.
Correct Answer:
A
— 4A
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Q. In a circuit with a 24V battery and two resistors in series (4Ω and 8Ω), what is the voltage drop across the 8Ω resistor?
A.
8V
B.
12V
C.
16V
D.
20V
Show solution
Solution
Total resistance R = 4Ω + 8Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 8Ω = I * R = 2A * 8Ω = 16V.
Correct Answer:
B
— 12V
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Q. In a circuit with a 24V battery and two resistors in series (8Ω and 4Ω), what is the voltage drop across the 4Ω resistor?
A.
6 V
B.
8 V
C.
12 V
D.
4 V
Show solution
Solution
Total resistance = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop across 4Ω = I * R = 2A * 4Ω = 8 V.
Correct Answer:
A
— 6 V
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Q. In a circuit with a 24V power supply and two resistors of 8Ω and 4Ω in series, what is the voltage drop across the 4Ω resistor?
A.
4.8 V
B.
8 V
C.
12 V
D.
16 V
Show solution
Solution
Total resistance R = 8Ω + 4Ω = 12Ω. Current I = V/R = 24V / 12Ω = 2A. Voltage drop V_R = I * R = 2A * 4Ω = 8 V.
Correct Answer:
C
— 12 V
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Q. In a discharging RC circuit, what is the expression for the voltage across the capacitor at time t?
A.
V = V0*e^(-t/RC)
B.
V = V0*(1 - e^(-t/RC))
C.
V = V0*t
D.
V = V0/t
Show solution
Solution
In a discharging RC circuit, the voltage across the capacitor at time t is given by V = V0*e^(-t/RC), where V0 is the initial voltage.
Correct Answer:
A
— V = V0*e^(-t/RC)
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Q. In a double-slit experiment, what causes the interference pattern?
A.
Reflection
B.
Refraction
C.
Diffraction
D.
Interference
Show solution
Solution
The interference pattern is caused by the superposition of light waves from the two slits.
Correct Answer:
D
— Interference
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Q. In a frictionless environment, if a 4 kg object is pushed with a force of 16 N, what will be its acceleration?
A.
2 m/s²
B.
4 m/s²
C.
6 m/s²
D.
8 m/s²
Show solution
Solution
Acceleration a = F/m = 16 N / 4 kg = 4 m/s².
Correct Answer:
B
— 4 m/s²
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Q. In a heat engine, if the input heat is 1000 J and the work done is 300 J, what is the efficiency of the engine?
A.
30%
B.
70%
C.
50%
D.
10%
Show solution
Solution
Efficiency is calculated as η = (Work output / Heat input) * 100%. Here, η = (300 J / 1000 J) * 100% = 30%.
Correct Answer:
A
— 30%
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Q. In a heat engine, if the input heat is 800 J and the output work is 300 J, what is the efficiency?
A.
0.375
B.
0.5
C.
0.25
D.
0.625
Show solution
Solution
Efficiency = Work output / Heat input = 300 J / 800 J = 0.375 or 37.5%.
Correct Answer:
A
— 0.375
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Q. In a heat engine, what is the purpose of the heat reservoir?
A.
To absorb heat
B.
To release heat
C.
To convert heat to work
D.
To increase efficiency
Show solution
Solution
The heat reservoir serves as a source of thermal energy that the heat engine absorbs heat from to perform work.
Correct Answer:
A
— To absorb heat
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Q. In a heat engine, what is the purpose of the heat sink?
A.
To absorb heat from the engine
B.
To release heat to the environment
C.
To increase efficiency
D.
To store energy
Show solution
Solution
The heat sink in a heat engine is used to release excess heat to the environment, allowing the engine to operate efficiently.
Correct Answer:
B
— To release heat to the environment
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Q. In a heat engine, what is the work done equal to?
A.
Heat input minus heat output
B.
Heat output minus heat input
C.
Total heat energy
D.
Change in internal energy
Show solution
Solution
The work done by a heat engine is equal to the heat input minus the heat output.
Correct Answer:
A
— Heat input minus heat output
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Q. In a p-n junction diode, what does the 'p' stand for?
A.
Positive charge carriers
B.
Photons
C.
Protons
D.
Potential difference
Show solution
Solution
The 'p' in p-n junction diode stands for positive charge carriers, which are holes created by the absence of electrons.
Correct Answer:
A
— Positive charge carriers
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Q. In a p-n junction diode, what type of charge carriers are predominant in the p-type region?
A.
Electrons
B.
Holes
C.
Ions
D.
Neutrons
Show solution
Solution
In the p-type region of a p-n junction diode, holes are the predominant charge carriers, created by the addition of acceptor impurities.
Correct Answer:
B
— Holes
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Q. In a parallel circuit with a 12 V battery and two resistors of 4 Ω and 6 Ω, what is the total current supplied by the battery?
A.
2 A
B.
3 A
C.
4 A
D.
1 A
Show solution
Solution
Total resistance, R_eq = 1/(1/4 + 1/6) = 2.4 Ω. Total current, I = V/R_eq = 12 V / 2.4 Ω = 5 A.
Correct Answer:
B
— 3 A
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Q. In a parallel circuit with a 12 V battery, what is the voltage across each branch?
A.
12 V
B.
6 V
C.
24 V
D.
0 V
Show solution
Solution
In a parallel circuit, the voltage across each branch is equal to the source voltage, which is 12 V.
Correct Answer:
A
— 12 V
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Q. In a parallel circuit with a 12 V supply, what is the voltage across each resistor?
A.
12 V
B.
6 V
C.
24 V
D.
0 V
Show solution
Solution
In a parallel circuit, the voltage across each component is equal to the supply voltage, so it is 12 V.
Correct Answer:
A
— 12 V
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Q. In a parallel circuit with a 12V battery and two resistors (4Ω and 12Ω), what is the total current supplied by the battery?
A.
1 A
B.
2 A
C.
3 A
D.
4 A
Show solution
Solution
Total current I = V/R_total. R_total = 1/(1/4 + 1/12) = 3Ω. I = 12V / 3Ω = 4 A.
Correct Answer:
C
— 3 A
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Q. In a parallel circuit with a total voltage of 12 V, what is the voltage across each resistor?
A.
6 V
B.
12 V
C.
24 V
D.
0 V
Show solution
Solution
In a parallel circuit, the voltage across each resistor is equal to the total voltage, so it is 12 V.
Correct Answer:
B
— 12 V
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Q. In a parallel circuit with three resistors of 2Ω, 3Ω, and 6Ω, what is the total resistance?
Show solution
Solution
1/Rtotal = 1/R1 + 1/R2 + 1/R3 = 1/2 + 1/3 + 1/6 = 1Ω.
Correct Answer:
A
— 1Ω
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Q. In a parallel circuit with two resistors of 6 ohms and 12 ohms, what is the equivalent resistance?
A.
4 ohms
B.
8 ohms
C.
2 ohms
D.
3 ohms
Show solution
Solution
For parallel resistors, 1/R_eq = 1/R1 + 1/R2. Here, 1/R_eq = 1/6 + 1/12 = 2/12 + 1/12 = 3/12, so R_eq = 4 ohms.
Correct Answer:
A
— 4 ohms
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Q. In a parallel circuit with two resistors of 6Ω and 12Ω, what is the total resistance?
Show solution
Solution
1/R_eq = 1/R1 + 1/R2 = 1/6 + 1/12 = 1/4. Therefore, R_eq = 4Ω.
Correct Answer:
A
— 4Ω
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